Derivative of an inverse function: I dx dy dy = dx IMPLICIT DIFFERENTIATION • GIVEN AN EQUATION INVOLVING FUNCTION OF x AND" TO FIND: dlY lX I.. Write quantity to be optimized as a
Trang 1REVIEW OF BASIC CALCULUS FOR BUSINESS, BI LOGY & PSYCHOLOGY MAJORS
• lim/(x) = L if/(x) is close to L for all x
x
-sufficiently close (but not equal) to a
•/(x) is continuous at x = a if:
l.f(a) is defined,
2 Iim/(x)=L exists, and
x 0
3 L=f(a)
INTEGRALS
THE DEFINITE INTEGRAL
• ETlex) BE CONTINUOUS ON la, hi
I Riemann Sum Definition of Definite
Integral
a Divide la, hi into n equal subintervals
of length h=b~a
b Let Xo = a, XI ' x 2, , x" = h denote the
endpoints of the subintervals They are
founu by: Xo = a, XI = a + Jr, X 2 = a +
2h, X3 = a + 3h, , x" = a + nh = h
c.Let m l , "'2 ' .•• m" denote the midpoints
of the subintervals They are found by:
m l =0.5(xo +XI)' m2 =0.5(xl +x 2) 1n3
= 0.5(X2 +x 3) "', mIl = 0.5(x" -1 +x,,).
.b
LI(x)dx:::::hlf(m l )+f(m2)+'" +/(m,,)I
b
2 Midpoint Rule: {I(x)dx =
"0
limhl/(xl ) +/(x 2) + +/(x,,) I
"
(h "
3 Trapezoid Rule: -",I(x)dx::::: 21/(xO) +
2/(xl ) + 2/(X2) + + 2/(x,,_I) +/(x,,)I
[ " h
4 Simpson's Rule: ."/(x)dx::::: 31 /(xo)+
4/(ml ) + 2/(xl ) + 4/(m 2) + 2/(x 2) + +
2/(x,,_I) + 4/(m,,) +j(x,,>1
THE INDEFINITE INTEGRAL
• F(x) IS CALLED AN ANTIDERIVATIVE
OF/(.i) , iF P(x) = lex)
I The most general anti derivative is
denoted V(x)dx
2 V(x)dx is also called the Indefinite
Integral oflex)
• Fundamental Theorem of Calculus
I If P(x) =f(x) and/(x) is continuous on
b
[a , h], then [I(x)dx = F(h) - F(a)
· 0
I II/(x) +g(x)ldx = V(x)dx + fg(x)dx
2 Ik/(x)dx = kff(x)dx if k is a constant
3 u"du= ~+I +C
4 I ~du =Inlul+C
5 Ieudu = eU+C
"
6.1fy =lex) ~ 0 on la, hI, [I(x)dx gives
· 0
the area under the curve b
7.lfJ(x) ~g(x) on la, hI, 1 [f(x)- g(x)ltL\"
gives the area between °the two curves
y =lex) and y = g(x)
8 Average value of f(x) on la, hi is I
b-a f."I(x)dx
9 Volum'e of the solid of revolution obtained by revolving about the x-axis
the region under the curve y = lex) from
h
x=atox=his [ll[/(x)fdx
·11
INTEGRATION BY PARTS
I Factor the integrand into two Parts:u and dv
2 Find du and v = f dv 3.Find f vdu
4 Set Judv = uv - Jvdu
INTEGRATION BY SUBSTITUTION
• TO SOLVE V(g(x»g'(x)dx
I Set u = g(x) , where g(x) is chosen so as
to simplity the integrand
2 Substitute u = g(x) and (Iu = g'(x)tL\" into the integrand
a This step usually requires multiplying
or dividing by a constant
3 Solve V(II)du = F(u) + C
4 Substitute II = g(x) to get the answer:
F(g(x»+c
IMPROPER INTEGRALS
• INFINlTE LIMITS OF INTEGRATION
I f.' I(x)dx= lim 1.f(x)dx (J
.h " -00 a h
2 L oo/(x)dX= !i"!.~/(x)dx
RIGHT ENDPOINTS
I If lex) is discontinuous at x = h
" h [I(x)dx = limj I(x)dx
• (l h - II 11
2 If lex) is discontinuous at x = a
1.,,1(x)dx = 1!~"!j"/(x)dx
APPLICATIONS DERIVATIVE BASICS
• DEFINITION OF DERIVATIVE
I /,(a) = Iim/(a+/~-1(0)
h ·0
2 If Y = f(x), thc derivative /'(x) is also denoted ~~
• FORMULAS:
I Power Rule: dd (x") = "X,,·I
2 dx(e"-'") = kekx
3 :/x(/1L\') = ~
4 General Power Rule:
d dx(I/(x)I") =1l1/(x)I,,·I/,(x)
5 dd lef(x») = ef( " "Ij'(x)
x
6 dx Iin/(x)] = I(x)
:/xlf(X) ± I:(x)I = /'(x) ±g'(x) ,
8 Constant Multiple Rule: "
4 11if(x)] = k/,(x) II
:fx I/(x)g(x)I /'(x)g(x) + j(x)g'(x)
] O.Quotient Rule:
4 [ I(X)j= f'(x)g(x)- I(x)g'(x)
dx g(x) Lg(x)] 2
I I.Chain Rule:
dx l/ W'\"))] =j'(g'\"» )g" '\"), or dx = du • dx
12 Derivative of an inverse function:
I
dx dy
dy = dx
IMPLICIT DIFFERENTIATION
• GIVEN AN EQUATION INVOLVING FUNCTION OF x AND)" TO FIND: dlY
lX
I DilTerentiatc both sides of the equation with respect to x treatingy as a function of C
X and applying the chain rule to each term involvingy (i.e ft: I/(y)] =/,(y) ~~) ~
2 Move all terms with ~~~ to left sidl: and ; all other terms to the right
3 Solve for ~~
Trang 2• STEPS TO FOLLOW IN SKETCHING
THE CURVE y = lex):
I Determine the domain of/ex)
2 Analyze all points where lex) is
discontinuous Sketch the graph near all
such points
3 Test for vertical, horizontal and oblique
asymptotes
a.f(x) has a vertical asymptote atx = a if:
limJ(x) = ±co or lim/(x) = ±co
b.f= (x) has a horizontal asymptotey = h if:
lim/(x) = h or lim lex) = h
c Sketch any asymptotes
4 Findl'(x) and/"(x)
5 Find all critical points These are points
x=a where I'(a) does not exist or
/,(a)=O Repeat steps 5.a & 5.b for
each critical point x = a:
a Iflex) is continuous at x = a
i./(x) has a relative maximum at x = a if:
(a) /,(a) = 0 and/"(a) < 0, or
(b) /'(x) > 0 to the left of a and
/'(x)<O to the right of a
ii,f(x) has a relative minimum atx=a if:
(a) /,(a) = 0 and/"(a) > 0, or
(b) /'(x) < 0 to the left of a and
/'(x) > 0 to the right of
b.Sketch/(x) near
6 Find all possible inflection points These
are points x = a where /"(x) does not
exist or/"(x) = O Repeat steps 6a & 6b
for each such x = a:
a lex) has an inflection point at x = a if
lex) is continuous at x = a and
i.f"(x) < 0 to the left ofa and/"(x) > 0
to the right ofa, or
ii /"(x) > 0 to the left ofa and/"(x) < 0
to the right of
b.Sketch/(x) near
7 If possible, plot the x-and y-intercepts
8 Finish the sketch
OPTIMIZATION PROBLEMS
• TO OPTIMIZE SOME QUANTITY
SUBJECT TO SOME CONSTRAINT:
I Identify and label quantity to be
maximized or minimized
2 Identify and label all other quantities
3 Write quantity to be optimized as a
function of the other variables This is
called the objective function (or
objective equation)
4 I f the objective function is a function of
more than one variable, find a constraint
equation relating the other variables
5 Use the constraint equation to write the
objective function as a function of only
one variable
6 Using the curve sketching techniques,
locate the maximum or minimum of the
objective function
• ~&I"'4~
• LETy=/(x)ANDASSUME/,(a) EX]STS
I The Equation of the Tangent Line to y =/(x)
at the point (a,/(a» is y - lea) =/,(a)(x-o)
2 The differential ofy is dy = /,(x)dx
3 Linear Approximation, or Approximation
by Differentials Set dx = ~x = x - a,
~y = lex) - lea)
~y = /,(a)~x = /,(a)dx If ~x is small then ~y:::::
That is,/(x) ::::: lea) +
4 The nth Taylor polynomial of/ex) centered
at x = a IS Pn(x) =lea) + l! +
j"(0)(X-O)2 j ( I ) (O)(X-O)"
MOTION
• FORMULA
If s = S(I) represents the pOSItIOn of an object at time I relative to some fixed point, then V(/)=s'(/) = velocity at time I and
a(/) = v'(I) = S"(/) = acceleration at time I
COST, REVENUE & PROFIT
I C(x) = cost ofproducingx units ofa product
2 P = p(x) = price per unit; (p = p(x) is also called the demand equation)
3 R(x) =xp = revenue made by producing
x units
4 P(x) = R(x) - C(x) = profit made by producing x units
5 C(x) = marginal cost
6 R '(x) = marginal revenue
7 r(x) = marginal profit
COMPOUNDING INTEREST
I If the interest is compounded for I years with m periods per year at the interest rate of r per annum, the compounded amount is: P = Po(1 + :;')ml
2 If interest is continuously compounded,
m~ co and the formula becomes:
P = lim P o(1 + :;')mt = P = Poe rt
m ""
3 The formula P = Poe rt gives the value at the end of t years, assuming continuously compounded interest Po is called the present value ofP to be received in I years and is given by the formula Po = Pe- rt
2
ELASTICITY OF DEMAND
• SOLVING FOR X IN THE DEMAND EQUATION P =p(x) GIVES x =/(p)
I Demand function which gives the quantity demanded x as a function of the price p
2 The elasticity of demand is:
£(P) = -p/'(p)
j(p)
• DEMAND IS ELASTIC AT P = pO IF
£(pO) > 1
In this case, an increase in price corresponds to a decrease in revenue
• DEMAND IS INELASTIC AT P = pO IF
£(pO) < 1
In this case, an increase in price corresponds to an increase in revenue
CONSUMERS'SURPLUS
• IF A COMMODITY HAS DEMAND EQUATION P = p(x)
Consumers' Surplus is given by
fa Ip(x) - p(a)ldx where a is the quantity
J demanded andp(a)
EXPONENTIAL GROWTH
• EXPONENTIAL GROWTH: y = Poe'"
I Satisfies the di tfcrential equation y' =ky
2 Po is the initial size, k > 0 is called the
growth constant
3 The time it takes for the size to double is given by: l~2
• EXPONENTIAL DECAY: y = Poe- A t
I Satisfies the differential equation y' =-"- y
2 Po is the initial size, A > 0 is called the
decay constant
3 The half life 11/2 is the time it takes for y
to become Po/2 It is found by 11/2 =
~l = 112
OTHER GROWTH CURVES
• THE LEARNING CURVE:y =M(1 e- k/ )
Satisfies the differential equation y' = k(M-y) , yeO) = 0 where M and k are
positive constants
• THE LOGISTIC GROWTH CURVE:
l+Be
equation y' = ky(M-y) where H, M and k
are positive constants
Trang 3For the continuous random variable X is a
function p(x) satistying: And p(x) ~ 0 if
H
·A
assume the values ofx lie in lA, BI
• THE PROBABILITY THAT
."
a :::X::: h is Pia :::X::: hi = I p(x)dx
."
j
• EXPECTED VALUE, OR MEAN OF X
.IJ
Given by m = E(X) = A xp(x)dx
• VARIANCE OF X: Given by 0 2 = var(X)=
l (x - Il)2P(x)dx = l x2p(x)dx - 112
COMMON PROBABILITY
DENSITY FUNCTIONS
• UNIFORM DISTRIBUTION FUNCTION:
p(x)=B~A ' Il=E(X)=BiA , var(X) (B~2A)2
• EXPONENTIAL DENSITY FUNCTION:
p(x) = Ae-Ax ]n this case, A = 0 B = 00,
J1 = E(X) = IIA, var(X) = IIA2
• NORMAL DENSITY FUNCTION:
with E(X) = J1 and vor (X) = 0 2 is:
_ (x-PI '
l2iia
[(X-JlV]
I2iiI a exp 2S2
PARTIAL DERIVATIVES
• WHERE f (x, y ) IS A FUNCTION OF
TWO VARIABLES x AND Y
I ~~ is the derivative of f ( x, y ) with
respe t to x, treating y as a constant
2.~ is the derivative of f(x ,y ) with
respect to y, treating x as a constant
3 a2
= ~ _ ~ is the second partial
a x UY o X
derivative of f(x ,y ) with respec t to x
twice, keepingy constant each time
a2 I a a f · h d I
axay = oy oX IS t e secon partla
derivative off( x ,y), first with respect
to x keepi ng y constant, then with
respect to y keeping x constant
5 Other notation for partial derivatives:
fx(X ,y) - ax , f.:J x,y) - a;'i, h y (X ,y) - ayax
• If f = f(x, y )
I d f = ax dx + ay dy = /., (x, y ) dx +J;,(x, y ) d y
2 Setting dx = ~x =x -a, d y = ~y = y - h
and ~f =f( x, y) -f(a , ), if ~xand ~ya r e
both small, then ~ f ;:::: df That is :
f (x , y ) ;:::: f ( a h ) +/., ( 0 h)~ x +.I;( a h )~y
RELATIVE EXTREMA TEST
• O LOCATE RELATIVE MAXIMA, RELATIVE MINIMA AND SADDLE
POINTS ON THE GRAPH OF z = f( x, y )
O For e ch ordered pair (0 , h) such that
rx (a, b) = 0 and ~( a,b) = 0 apply the
a
following tes
2 f a2
2 Set A = - 2( a b) , B = - 2 ( a b),
c = aa xay f ( ab , )a ndD=AB - C2,
a lf D > 0 and A > 0, then f ( x , y) has a
relative minimum at (a, h )
b lf D > 0 and A < 0, then f( x ,y ) has a
relative maximum at ( a, h)
c If D < 0, t h e n f (x ,y) has a saddle point
at (a , h )
d.f D = 0 then the test fails /(x,y) may
or may not have an extremum or saddle point at (a, h)
THE METHOD OF LAGRANGE MULTIPLIERS
• SOLVES CONSTRAINED OPTIMIZATION
PROBLEMS TO MAXIMIZE OR MINIM IZE f( x ,y) SUBJECT TO THE
CONSTRAINT g(x, y)=O
I Define the new function
F( x, y , A )= f ( x, y) +A g ( x ,y )
2 Solve the system of three equatons:
aF
a ax =0,
a F
b.ay=O, and
c ~~=0 simultaneously
This is usually accomplished in four steps:
Step I: Solve a and b for Aand equate
the solutio s Step 2: Solve the resulting equation for
one of the variables, x or y
Step 3: Substitute this expre sion for x
or y into equation c and solve the
resulting equation of one variable for
the other variable
Step 4: Substitute the value found in Step
3 into the eq uation foun d in Step 2 Use
one ofthe equations from Step I to fi nd
A This gives the value of x and y
3
I If R is the region in the plane bounded by
the two curvesy =g (x),y =" (x) and the two
vertical lines x = 0, x = h, then the d uble
integral IfI(x, y ) dxdy is equal to the
iterated integral.{ ~ (.\./(x,y)dy dx
2 To evaluate the iterated integral
1 = t( ( dxf (X, Y )dY ) d x
'1/ • g (x)
a find an antiderivative F (x,y) for f(x,y )
with respect to y keeping x constant
That is: t~ = f(x , y )
b.Set: 1= [[F( x ,h( x »
'1/
c Solve this integral The integrand is a function of o e variable
• A DIFFERENTIAL EQUATION IS: Any equation involving a derivative For example, it could be an equation involving
~~ (or y', or y'(x», y and x
• A SOLUTION IS: A function y = y(x), such that ~~, y and x satisty the original equation
• AN INITIAL VALUE PROBLEM al so specifies the value of the solution yea) at some point x = a
• SIMPLE DIFFERENTIAL EQUATIONS can be solved by separation of variables and integration For example, the equation
f(x) = g(y) ~~ can be written as f(x)dx = g(y)dy and can be solved by integrating
f· f ' dy
both sides: I(x)= g(y) dx'
FORMULAS FROM PRE-CALCULUS
LOGARITHMS
I.y = Inx if and only if x = eY=
2
3 e1n x = x
4 e · \"eY = eX+.r
x
5 e e (-)'
Y = e'
6 (ex)y = e'\)'
7 eO =
8.ln(xy) = In x +
9.ln(x/y) =
IO.ln(x.l') = y In x 11.1n I = 0
12.lne=1
4
Trang 4ALGEBRAIC FORMULAS
~ 1.I f a # 0, the solutions to ax2 + bx + c = 0
Z are gIven by x = 2a
III 2.Point-slope equation ofa line:
Y - Yo = m(x -xo)·
= DEVELOP A PROBLEM SOLVING STRATEGY
~
• Three key issues
1 Understand the business or scientific
principle required to solve the problem
2.Develop a correct mathematical strategy
3.Logically approach solving the problem
• Eight useful steps in problem solving:
I Prepare a rough sketch or ~ ,lI
diagram based on the subject J., $
of the problem For business
applications, sct this up using business
terms; for science, use physical variables
2 Identify all relevant variables, concepts
and constants
Note: Do not simply search for the
"right" equation in your notes or text
You may have to select your own
variables to solve the problem
3.Describe the problem using appropriate
mathematical relationships or graphs
4.0btain any constants from the stated
problem or textbook Make sure you
have all the essential data S
Hint: You may have extra JlJl
5.The hard part: Derive a mathematical
expression for the problem Make sure
that the equation, constants and data give
the right unit for the final answer
6 Carry out the appropriate mathematical
manipulation, differentiate, integrate,
find limits, etc
7.The easy part: Plug numbers into the
equation Obtain a quick approximate
answer, then use a calculator to _I
obtain an exact numerical answer 'J
8.Check the final answer using the
original statement of the problem, your
sketch and common sense; are the units
correct? Sign? Magnitude?
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ISBN-13 978-157222841-2
ISBN-10· 157222841-5
= ~
911~lllJll~~llllllI1 1!IJI111 11 1 111Irllillll
Perimeter: The perimeter, P, of a two-dimensional shape is the sum of all side lengths Area: The area, A, of a two dimensional shape is the number of squan: units that can be put region enclosed by the sides Note: Area is obtained through some combination of
heights and bases , which always form 90" angles with each other, except ill circles
Volume: The volume, V , of a three-dimensional shape is the number of cubic units that put in the region enclosed by all the sides
Square Area:
A = b 2 ; If h = 8 then: A = 64 square
Rectangle Ar
A = hb , or A = [w; If h = 4 and b = 12, then
A = (4)(1 2), A = 48 square units
Triangle Area:
A = 1bh; If h = 8 and b = 12, then:
A = 1 (8)(12), A = 48 square units
A = hb; If h = 6 and b =
A = (6)(9), A = 54 square units
Trapezoid Area:
A = 1"(bl + b2); If" = 9 bl = 8 and b2 = 12, then:
A = 1(9)(8 + 12), A = 1(9)(20), A = 90 square units
Circle Area
A = nr 2 If r = 5 the
A = n(5)2 = (3.1 4)25 = 78.5 square
Circumference: C = 21tr; If r = 5 the
If a right triangle has hypotenuse c and sides Q and b, then: (J = 0 2 + b 2
u
V = [wh; If [= 12, HI = 3 and h =
V = e3 ; each edge length, e, is equal to the {1O Ife = 8, then: V= (8)(8)(8), V= 512 cubic units e e
Cylinde
V = n r 2 h If radius r = 9 and" = 8 then
r :· "
~
V= 1 n(6)2(8), V= 1 (3.14)(36)(8), V= 301.44 cubic units
Triangular Prism Volume:
V= (area oftriangle)h; If 12 has an area equal to 1(5)(12), then : 4
V = 30h and if h = 8, then: V = (30)(8), V = 240 eubic units ~
Rectangular Pyramid Volume:
V = 1(area of rectangle)"; If [ = 5 and w = 4 the rectangle r
"
has an area of20, then: V= 1(20)" and if" = 9 then:
V = 1(20)(9), V = 60 cubic units
Sphere Vol
V = ~ n r3; If radius r = 5 then: V = ~ ( 3 1 4)(5) 3, V = 523.3 cubic unit